Kinetic Energy Related to Potential

AI Thread Summary
The discussion centers on determining the height, h1, at which kinetic energy (K) equals twice the potential energy (U) for an object dropped from a height h0. The relationship between kinetic and potential energy is established with the equations K = 2U and the total energy equation ET = K - U. It is clarified that before the drop, the total energy is E = mgh, and after falling, it becomes E' = 3U'. The conclusion drawn is that the new height h' is one-third of the original height h0, or h' = 1/3 h0. This analysis highlights the interplay between kinetic and potential energy in a gravitational field.
Noms
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Object dropped from rest at h0meters. At what height, h1, will K=2U? g=9.8m/s2K=2U
ET=K-U
(.5)mv2=2(mgh)
v2=gh
(v2)/g=h


Thanks. :)
 
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Noms said:
Object dropped from rest at h0meters. At what height, h1, will K=2U? g=9.8m/s2


K=2U
ET=K-U



(.5)mv2=2(mgh)
v2=gh
(v2)/g=h


Thanks. :)
Your question does not make sense. Please give us the entire question exactly as you were given it.

AM
 
Before dropping the object the total energy is E = K + U = 0+mgh.

After falling, the total energy will be E' = K' + U' = 2U' + U' = 3U'

As we know, E = E'

Then, mgh = 3mgh' --> h' = 1/3 h
 
AM- Sorry I wasn't clear enough for you, but thanks for looking it over,

M- That's just what I needed. :)
 
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